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nevsk [136]
2 years ago
7

Can somebody help me on this thanks

Mathematics
1 answer:
sukhopar [10]2 years ago
5 0

Answer:

Add 3/4 to both sides (the last option)

Step-by-step explanation:

Since you want to isolate f and the expression on one side is f - 3/4, you have to add 3/4 so that the -3/4 cancels out. When you do that, you also have to add 3/4 to the other side so that the equation stays balanced.

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No. They are not in proportion

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\sf \dfrac{4}{5} \ \text{is in the simplest form}

\sf \text{Simplest form of $\dfrac{12}{20}$} = \dfrac{12 \div 4}{20 \div 4}=\dfrac{3}{5}

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Please Help! This is a trigonometry question.
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\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}


\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}

\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}

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\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}

________


\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}

\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}


<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2150237


\large\textsf{I hope it helps.}


Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>

</span>
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Answer:

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Step-by-step explanation:

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